Re: [math-fun] something linguistic
with thanks to Hans: similar pronounciation or perfect match (pron. & orthography = ****) ware :: where way / weigh :: whey we'll / weal :: wheel we'r :: whir weald / wield :: wheeled weather / wether :: whether **** wen :: when **** wet :: whet **** witch :: which wig :: whig (??) **** wile :: while **** wiled :: whiled **** wine :: whine **** word :: whirred wit :: whit **** wither :: whither **** wye :: why cute language you have, difficult to write it is! Wouter. -----Original Message----- From: Hans Havermann Sent: Friday, September 18, 2015 4:56 PM To: Wouter Meeussen Subject: Re: [math-fun] something linguistic http://www.cooper.com/alan/homonym_list.html Scroll down to the W.
What about we, whee ? On 18-Sep-15 13:02, Wouter Meeussen wrote:
with thanks to Hans: similar pronounciation or perfect match (pron. & orthography = ****) ware :: where way / weigh :: whey we'll / weal :: wheel we'r :: whir weald / wield :: wheeled weather / wether :: whether **** wen :: when **** wet :: whet **** witch :: which wig :: whig (??) **** wile :: while **** wiled :: whiled **** wine :: whine **** word :: whirred wit :: whit **** wither :: whither **** wye :: why
cute language you have, difficult to write it is!
Wouter.
-----Original Message----- From: Hans Havermann Sent: Friday, September 18, 2015 4:56 PM To: Wouter Meeussen Subject: Re: [math-fun] something linguistic http://www.cooper.com/alan/homonym_list.html
Scroll down to the W.
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A century ago, careful English speakers would pronounce ``whee'' as ``hwee'' with a noticeable aitch. Someone mentioned this in connexion with ``some American dialects''. I suspect that, down the years, the aitch has largely disappeared. Ditto for (almost?) all `wh' words? R. On Fri, 18 Sep 2015, Dan Asimov wrote:
What about we, wee, whee, all the way home?
—Dan
On Sep 18, 2015, at 10:22 AM, Mike Speciner <ms@alum.mit.edu> wrote:
What about we, whee ?
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I still pronounce the h when I'm enunciating. On Fri, Sep 18, 2015 at 10:38 AM, rkg <rkg@ucalgary.ca> wrote:
A century ago, careful English speakers would pronounce ``whee'' as ``hwee'' with a noticeable aitch. Someone mentioned this in connexion with ``some American dialects''. I suspect that, down the years, the aitch has largely disappeared. Ditto for (almost?) all `wh' words? R.
On Fri, 18 Sep 2015, Dan Asimov wrote:
What about we, wee, whee, all the way home?
—Dan
On Sep 18, 2015, at 10:22 AM, Mike Speciner <ms@alum.mit.edu> wrote:
What about we, whee ?
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-- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
Many people still say the HW. Except for "who". Did people ever say HWOO ? (Other than hwen imitating owls?) —Dan
On Sep 18, 2015, at 10:38 AM, rkg <rkg@ucalgary.ca> wrote:
A century ago, careful English speakers would pronounce ``whee'' as ``hwee'' with a noticeable aitch. Someone mentioned this in connexion with ``some American dialects''. I suspect that, down the years, the aitch has largely disappeared. Ditto for (almost?) all `wh' words? R.
On Fri, 18 Sep 2015, Dan Asimov wrote:
What about we, wee, whee, all the way home?
On Sep 18, 2015, at 10:22 AM, Mike Speciner <ms@alum.mit.edu <mailto:ms@alum.mit.edu>> wrote:
What about we, whee ?
I've just remembered Flanders & Swann's ``I'm a g'nu!'' ``... and nobody knows wuhoo's wuhoo''. `` nor am I in the least like that dreadful hearty beast; oh g'no, g'no, g'no, I'm a g'nu!'' R. On Fri, 18 Sep 2015, rkg wrote:
A century ago, careful English speakers would pronounce ``whee'' as ``hwee'' with a noticeable aitch. Someone mentioned this in connexion with ``some American dialects''. I suspect that, down the years, the aitch has largely disappeared. Ditto for (almost?) all `wh' words? R.
On Fri, 18 Sep 2015, Dan Asimov wrote:
What about we, wee, whee, all the way home?
—Dan
On Sep 18, 2015, at 10:22 AM, Mike Speciner <ms@alum.mit.edu> wrote:
What about we, whee ?
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For some reason, it's "Have some madeira, m'dear" that sticks in my mind. —Dan
On Sep 18, 2015, at 10:47 AM, rkg <rkg@ucalgary.ca> wrote:
I've just remembered Flanders & Swann's ``I'm a g'nu!''
``... and nobody knows wuhoo's wuhoo''.
`` nor am I in the least like that dreadful hearty beast; oh g'no, g'no, g'no, I'm a g'nu!'' R.
On Fri, 18 Sep 2015, rkg wrote:
A century ago, careful English speakers would pronounce ``whee'' as ``hwee'' with a noticeable aitch. Someone mentioned this in connexion with ``some American dialects''. I suspect that, down the years, the aitch has largely disappeared. Ditto for (almost?) all `wh' words? R.
On Fri, 18 Sep 2015, Dan Asimov wrote:
What about we, wee, whee, all the way home?
—Dan
On Sep 18, 2015, at 10:22 AM, Mike Speciner <ms@alum.mit.edu> wrote: What about we, whee ?
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On 18/09/2015 18:47, rkg wrote:
I've just remembered Flanders & Swann's ``I'm a g'nu!''
``... and nobody knows wuhoo's wuhoo''.
`` nor am I in the least like that dreadful hearty beast; oh g'no, g'no, g'no, I'm a g'nu!'' R.
Hartebeest. (A kind of antelope.) It is true but not universally known that before F&S wrote that song, "gnu" was pretty much always pronounced "nu", but now the initial "g" is extremely common. -- g
On Sep 18, 2015, at 12:30 PM, Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
On 18/09/2015 18:47, rkg wrote:
I've just remembered Flanders & Swann's ``I'm a g'nu!''
``... and nobody knows wuhoo's wuhoo''.
`` nor am I in the least like that dreadful hearty beast; oh g'no, g'no, g'no, I'm a g'nu!'' R.
Hartebeest. (A kind of antelope.)
It is true but not universally known that before F&S wrote that song, "gnu" was pretty much always pronounced "nu", but now the initial "g" is extremely common.
I've heard the same thing. So I looked it up in (what I consider) the 6 most respectable dictionaries at http://onelook.com/ <http://onelook.com/>, as well as in the online OED, and for what it's worth every one of these lists only the NOO or NYOO pronunciations — none starting with the G sound. Maybe the initial G sound came on the coattails of Gnu, Not Unix. —Dan
The song of patriotic prejudice is my favorite Flanders and Swann composition. It's not that you're wicked or naturally bad // It's knowing you're foreign that's driving you mad. On Fri, Sep 18, 2015 at 1:49 PM, Dan Asimov <asimov@msri.org> wrote:
On Sep 18, 2015, at 12:30 PM, Gareth McCaughan < gareth.mccaughan@pobox.com> wrote:
On 18/09/2015 18:47, rkg wrote:
I've just remembered Flanders & Swann's ``I'm a g'nu!''
``... and nobody knows wuhoo's wuhoo''.
`` nor am I in the least like that dreadful hearty beast; oh g'no, g'no, g'no, I'm a g'nu!'' R.
Hartebeest. (A kind of antelope.)
It is true but not universally known that before F&S wrote that song, "gnu" was pretty much always pronounced "nu", but now the initial "g" is extremely common.
I've heard the same thing. So I looked it up in (what I consider) the 6 most respectable dictionaries at http://onelook.com/ <http://onelook.com/>, as well as in the online OED, and for what it's worth every one of these lists only the NOO or NYOO pronunciations — none starting with the G sound.
Maybe the initial G sound came on the coattails of Gnu, Not Unix.
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-- Thane Plambeck tplambeck@gmail.com http://counterwave.com/
My favourite is probably Misalliance, with the honeysuckle and the bindweed.
Sent: Saturday, September 19, 2015 at 12:52 AM From: "Thane Plambeck" <tplambeck@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] something linguistic
The song of patriotic prejudice is my favorite Flanders and Swann composition.
It's not that you're wicked or naturally bad // It's knowing you're foreign that's driving you mad.
On Fri, Sep 18, 2015 at 1:49 PM, Dan Asimov <asimov@msri.org> wrote:
On Sep 18, 2015, at 12:30 PM, Gareth McCaughan < gareth.mccaughan@pobox.com> wrote:
On 18/09/2015 18:47, rkg wrote:
I've just remembered Flanders & Swann's ``I'm a g'nu!''
``... and nobody knows wuhoo's wuhoo''.
`` nor am I in the least like that dreadful hearty beast; oh g'no, g'no, g'no, I'm a g'nu!'' R.
Hartebeest. (A kind of antelope.)
It is true but not universally known that before F&S wrote that song, "gnu" was pretty much always pronounced "nu", but now the initial "g" is extremely common.
I've heard the same thing. So I looked it up in (what I consider) the 6 most respectable dictionaries at http://onelook.com/ <http://onelook.com/>, as well as in the online OED, and for what it's worth every one of these lists only the NOO or NYOO pronunciations — none starting with the G sound.
Maybe the initial G sound came on the coattails of Gnu, Not Unix.
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On 19/09/2015 01:54, Adam P. Goucher wrote:
My favourite is probably Misalliance, with the honeysuckle and the bindweed.
Misalliance. Bedstead Men. The Slow Train. Not a big fan of Madeira (a clever, charming, witty song about rape) or for that matter Tonga (a clever, charming, witty song about rape). I fear we have gone some way astray from actual Fun About Math here, but perhaps it's worth mentioning that F&S wrote a song that states the Laws of Thermodynamics. Oh, and my wife maintains that "The Warthog" is really about mathematics students at university. Or, more generally, nerds. -- g
On Sat, 19 Sep 2015, Adam P. Goucher wrote:
My favourite is probably Misalliance, with the honeysuckle and the bindweed.
I probably shouldn't say this out loud, but I sing "Madeira, M'dear" to my wife frequently. -- Tom Duff. The Lessons of history teach us that no one learns the lessons that history teaches us.
Brought to mind by Wouter's question is: Consider the group generated by the letters A through Z, with the relations that any two English homonyms are equal. Prove that the groups is trivial. Wouter's question was to show h is trivial, so 25 letters to go :-). --Michael -- Forewarned is worth an octopus in the bush.
Michael, do you mean homophones (two words spelled differently that sound alike)? Homonyms are spelled the same. TYRE = TIRE, so Y=I, if that helps. On Fri, Sep 25, 2015 at 3:57 PM, Michael Kleber <michael.kleber@gmail.com> wrote:
Brought to mind by Wouter's question is: Consider the group generated by the letters A through Z, with the relations that any two English homonyms are equal. Prove that the groups is trivial.
Wouter's question was to show h is trivial, so 25 letters to go :-).
--Michael
-- Forewarned is worth an octopus in the bush. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
TOO = TO = TWO shows that W and O are trivial On Fri, Sep 25, 2015 at 4:23 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Michael, do you mean homophones (two words spelled differently that sound alike)? Homonyms are spelled the same.
TYRE = TIRE, so Y=I, if that helps.
On Fri, Sep 25, 2015 at 3:57 PM, Michael Kleber <michael.kleber@gmail.com> wrote:
Brought to mind by Wouter's question is: Consider the group generated by the letters A through Z, with the relations that any two English homonyms are equal. Prove that the groups is trivial.
Wouter's question was to show h is trivial, so 25 letters to go :-).
--Michael
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And I haven't seen the proof that H is trivial, but assuming that, then in those dialects that pronounce initial wh and initial w identically, WHETHER = WEATHER shows that A is trivial WHICH = WITCH shows that T is trivial. DAMN = DAM, so N is trivial On Fri, Sep 25, 2015 at 4:29 PM, Andy Latto <andy.latto@pobox.com> wrote:
TOO = TO = TWO shows that W and O are trivial
On Fri, Sep 25, 2015 at 4:23 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Michael, do you mean homophones (two words spelled differently that sound alike)? Homonyms are spelled the same.
TYRE = TIRE, so Y=I, if that helps.
On Fri, Sep 25, 2015 at 3:57 PM, Michael Kleber <michael.kleber@gmail.com> wrote:
Brought to mind by Wouter's question is: Consider the group generated by the letters A through Z, with the relations that any two English homonyms are equal. Prove that the groups is trivial.
Wouter's question was to show h is trivial, so 25 letters to go :-).
--Michael
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WHEN = WEN proves H unit. On Fri, Sep 25, 2015 at 4:32 PM, Andy Latto <andy.latto@pobox.com> wrote:
And I haven't seen the proof that H is trivial, but assuming that, then in those dialects that pronounce initial wh and initial w identically,
WHETHER = WEATHER shows that A is trivial
WHICH = WITCH shows that T is trivial.
DAMN = DAM, so N is trivial
On Fri, Sep 25, 2015 at 4:29 PM, Andy Latto <andy.latto@pobox.com> wrote:
TOO = TO = TWO shows that W and O are trivial
On Fri, Sep 25, 2015 at 4:23 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Michael, do you mean homophones (two words spelled differently that sound alike)? Homonyms are spelled the same.
TYRE = TIRE, so Y=I, if that helps.
On Fri, Sep 25, 2015 at 3:57 PM, Michael Kleber < michael.kleber@gmail.com> wrote:
Brought to mind by Wouter's question is: Consider the group generated by the letters A through Z, with the relations that any two English homonyms are equal. Prove that the groups is trivial.
Wouter's question was to show h is trivial, so 25 letters to go :-).
--Michael
-- Forewarned is worth an octopus in the bush. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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HOLEY = HOLY, therefore E = 1. On Fri, Sep 25, 2015 at 4:43 PM, Allan Wechsler <acwacw@gmail.com> wrote:
WHEN = WEN proves H unit.
On Fri, Sep 25, 2015 at 4:32 PM, Andy Latto <andy.latto@pobox.com> wrote:
And I haven't seen the proof that H is trivial, but assuming that, then in those dialects that pronounce initial wh and initial w identically,
WHETHER = WEATHER shows that A is trivial
WHICH = WITCH shows that T is trivial.
DAMN = DAM, so N is trivial
On Fri, Sep 25, 2015 at 4:29 PM, Andy Latto <andy.latto@pobox.com> wrote:
TOO = TO = TWO shows that W and O are trivial
On Fri, Sep 25, 2015 at 4:23 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Michael, do you mean homophones (two words spelled differently that sound alike)? Homonyms are spelled the same.
TYRE = TIRE, so Y=I, if that helps.
On Fri, Sep 25, 2015 at 3:57 PM, Michael Kleber < michael.kleber@gmail.com> wrote:
Brought to mind by Wouter's question is: Consider the group generated by the letters A through Z, with the relations that any two English homonyms are equal. Prove that the groups is trivial.
Wouter's question was to show h is trivial, so 25 letters to go :-).
--Michael
-- Forewarned is worth an octopus in the bush. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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And thus, since INDITE = INDICT, T = TE = CT, so C = 1. On Fri, Sep 25, 2015 at 4:55 PM, Allan Wechsler <acwacw@gmail.com> wrote:
HOLEY = HOLY, therefore E = 1.
On Fri, Sep 25, 2015 at 4:43 PM, Allan Wechsler <acwacw@gmail.com> wrote:
WHEN = WEN proves H unit.
On Fri, Sep 25, 2015 at 4:32 PM, Andy Latto <andy.latto@pobox.com> wrote:
And I haven't seen the proof that H is trivial, but assuming that, then in those dialects that pronounce initial wh and initial w identically,
WHETHER = WEATHER shows that A is trivial
WHICH = WITCH shows that T is trivial.
DAMN = DAM, so N is trivial
On Fri, Sep 25, 2015 at 4:29 PM, Andy Latto <andy.latto@pobox.com> wrote:
TOO = TO = TWO shows that W and O are trivial
On Fri, Sep 25, 2015 at 4:23 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Michael, do you mean homophones (two words spelled differently that sound alike)? Homonyms are spelled the same.
TYRE = TIRE, so Y=I, if that helps.
On Fri, Sep 25, 2015 at 3:57 PM, Michael Kleber < michael.kleber@gmail.com> wrote:
Brought to mind by Wouter's question is: Consider the group generated by the letters A through Z, with the relations that any two English homonyms are equal. Prove that the groups is trivial.
Wouter's question was to show h is trivial, so 25 letters to go :-).
--Michael
-- Forewarned is worth an octopus in the bush. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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I am guessing that the rest of the alphabet now falls like dominoes. On Fri, Sep 25, 2015 at 4:56 PM, Allan Wechsler <acwacw@gmail.com> wrote:
And thus, since INDITE = INDICT, T = TE = CT, so C = 1.
On Fri, Sep 25, 2015 at 4:55 PM, Allan Wechsler <acwacw@gmail.com> wrote:
HOLEY = HOLY, therefore E = 1.
On Fri, Sep 25, 2015 at 4:43 PM, Allan Wechsler <acwacw@gmail.com> wrote:
WHEN = WEN proves H unit.
On Fri, Sep 25, 2015 at 4:32 PM, Andy Latto <andy.latto@pobox.com> wrote:
And I haven't seen the proof that H is trivial, but assuming that, then in those dialects that pronounce initial wh and initial w identically,
WHETHER = WEATHER shows that A is trivial
WHICH = WITCH shows that T is trivial.
DAMN = DAM, so N is trivial
On Fri, Sep 25, 2015 at 4:29 PM, Andy Latto <andy.latto@pobox.com> wrote:
TOO = TO = TWO shows that W and O are trivial
On Fri, Sep 25, 2015 at 4:23 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Michael, do you mean homophones (two words spelled differently that sound alike)? Homonyms are spelled the same.
TYRE = TIRE, so Y=I, if that helps.
On Fri, Sep 25, 2015 at 3:57 PM, Michael Kleber < michael.kleber@gmail.com> wrote:
> Brought to mind by Wouter's question is: Consider the group generated by > the letters A through Z, with the relations that any two English homonyms > are equal. Prove that the groups is trivial. > > Wouter's question was to show h is trivial, so 25 letters to go :-). > > --Michael > > > -- > Forewarned is worth an octopus in the bush. > _______________________________________________ > math-fun mailing list > math-fun@mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Given that W is trivial (courtesy of Andy Latto), KNOW = NO establishes the triviality of K. CELL = SELL ==> C = S SCENT = CENT ==> S is trivial (and therefore so is C) ... Rarer letters are harder to destroy, but: CUE = QUEUE ==> UE = QUEUE ==> QUE is trivial We've already shown that E is an identity, so QU is trivial. U can be destroyed by FLOWER = FLOUR ==> WE = U, so Q is also trivial. At the moment, we're left with: _B_D_FG_IJKLM__P_R___V_XYZ KNIGHT = NIGHT easily gets rid of K, and we have Y = I = E (from TYRE = TIRE and COMPLIMENT = COMPLEMENT), so those all go: _B_D_FG__JKLM__P_R___V_X_Z GNOME = NOME (from elliptic functions) kills the G. _B_D_F___JKLM__P_R___V_X_Z Also, FAT = PHAT to deduce F = P, but I can't immediately trivialise either.
Sent: Friday, September 25, 2015 at 9:32 PM From: "Andy Latto" <andy.latto@pobox.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] something linguistic
And I haven't seen the proof that H is trivial, but assuming that, then in those dialects that pronounce initial wh and initial w identically,
WHETHER = WEATHER shows that A is trivial
WHICH = WITCH shows that T is trivial.
DAMN = DAM, so N is trivial
On Fri, Sep 25, 2015 at 4:29 PM, Andy Latto <andy.latto@pobox.com> wrote:
TOO = TO = TWO shows that W and O are trivial
On Fri, Sep 25, 2015 at 4:23 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Michael, do you mean homophones (two words spelled differently that sound alike)? Homonyms are spelled the same.
TYRE = TIRE, so Y=I, if that helps.
On Fri, Sep 25, 2015 at 3:57 PM, Michael Kleber <michael.kleber@gmail.com> wrote:
Brought to mind by Wouter's question is: Consider the group generated by the letters A through Z, with the relations that any two English homonyms are equal. Prove that the groups is trivial.
Wouter's question was to show h is trivial, so 25 letters to go :-).
--Michael
-- Forewarned is worth an octopus in the bush. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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TRIVIALIZE = TRIVIALISE (S=Z) On Fri, Sep 25, 2015 at 2:14 PM, Adam P. Goucher <apgoucher@gmx.com> wrote:
Given that W is trivial (courtesy of Andy Latto), KNOW = NO establishes the triviality of K.
CELL = SELL ==> C = S SCENT = CENT ==> S is trivial (and therefore so is C)
...
Rarer letters are harder to destroy, but:
CUE = QUEUE ==> UE = QUEUE ==> QUE is trivial
We've already shown that E is an identity, so QU is trivial.
U can be destroyed by FLOWER = FLOUR ==> WE = U, so Q is also trivial.
At the moment, we're left with:
_B_D_FG_IJKLM__P_R___V_XYZ
KNIGHT = NIGHT easily gets rid of K, and we have Y = I = E (from TYRE = TIRE and COMPLIMENT = COMPLEMENT), so those all go:
_B_D_FG__JKLM__P_R___V_X_Z
GNOME = NOME (from elliptic functions) kills the G.
_B_D_F___JKLM__P_R___V_X_Z
Also, FAT = PHAT to deduce F = P, but I can't immediately trivialise either.
Sent: Friday, September 25, 2015 at 9:32 PM From: "Andy Latto" <andy.latto@pobox.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] something linguistic
And I haven't seen the proof that H is trivial, but assuming that, then in those dialects that pronounce initial wh and initial w identically,
WHETHER = WEATHER shows that A is trivial
WHICH = WITCH shows that T is trivial.
DAMN = DAM, so N is trivial
On Fri, Sep 25, 2015 at 4:29 PM, Andy Latto <andy.latto@pobox.com> wrote:
TOO = TO = TWO shows that W and O are trivial
On Fri, Sep 25, 2015 at 4:23 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Michael, do you mean homophones (two words spelled differently that sound alike)? Homonyms are spelled the same.
TYRE = TIRE, so Y=I, if that helps.
On Fri, Sep 25, 2015 at 3:57 PM, Michael Kleber < michael.kleber@gmail.com> wrote:
Brought to mind by Wouter's question is: Consider the group generated by the letters A through Z, with the relations that any two English homonyms are equal. Prove that the groups is trivial.
Wouter's question was to show h is trivial, so 25 letters to go :-).
--Michael
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I forgot to cross out K. Also, GENES = JEANS eliminates J. _B_D_F_____LM__P_R___V_X_Z BARMY = BALMY proves L = R (the lallation relation!).
Sent: Friday, September 25, 2015 at 10:14 PM From: "Adam P. Goucher" <apgoucher@gmx.com> To: math-fun@mailman.xmission.com Subject: Re: [math-fun] something linguistic
Given that W is trivial (courtesy of Andy Latto), KNOW = NO establishes the triviality of K.
CELL = SELL ==> C = S SCENT = CENT ==> S is trivial (and therefore so is C)
...
Rarer letters are harder to destroy, but:
CUE = QUEUE ==> UE = QUEUE ==> QUE is trivial
We've already shown that E is an identity, so QU is trivial.
U can be destroyed by FLOWER = FLOUR ==> WE = U, so Q is also trivial.
At the moment, we're left with:
_B_D_FG_IJKLM__P_R___V_XYZ
KNIGHT = NIGHT easily gets rid of K, and we have Y = I = E (from TYRE = TIRE and COMPLIMENT = COMPLEMENT), so those all go:
_B_D_FG__JKLM__P_R___V_X_Z
GNOME = NOME (from elliptic functions) kills the G.
_B_D_F___JKLM__P_R___V_X_Z
Also, FAT = PHAT to deduce F = P, but I can't immediately trivialise either.
Sent: Friday, September 25, 2015 at 9:32 PM From: "Andy Latto" <andy.latto@pobox.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] something linguistic
And I haven't seen the proof that H is trivial, but assuming that, then in those dialects that pronounce initial wh and initial w identically,
WHETHER = WEATHER shows that A is trivial
WHICH = WITCH shows that T is trivial.
DAMN = DAM, so N is trivial
On Fri, Sep 25, 2015 at 4:29 PM, Andy Latto <andy.latto@pobox.com> wrote:
TOO = TO = TWO shows that W and O are trivial
On Fri, Sep 25, 2015 at 4:23 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Michael, do you mean homophones (two words spelled differently that sound alike)? Homonyms are spelled the same.
TYRE = TIRE, so Y=I, if that helps.
On Fri, Sep 25, 2015 at 3:57 PM, Michael Kleber <michael.kleber@gmail.com> wrote:
Brought to mind by Wouter's question is: Consider the group generated by the letters A through Z, with the relations that any two English homonyms are equal. Prove that the groups is trivial.
Wouter's question was to show h is trivial, so 25 letters to go :-).
--Michael
-- Forewarned is worth an octopus in the bush. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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-- Andy.Latto@pobox.com
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I missed Allan's find of PHAT = FAT, which gives us F, so we're down to R and V. I don't buy BARMY = BALMY. Merriam Webster online gives multiple pronunciations for BALMY, but none match BARMY. In general, there's no R sound in a lot of words containing an R in non-rhotic accents, but they generally involve a modification of the preceding vowel. I think we want some word where the R isn't pronounced even in a rhotic accent (or where the R sound exists in a homonym that does not contain the letter R). Alternatively, two homonyms, one with R and one with RR, might be easier to find. Andy On Fri, Sep 25, 2015 at 5:35 PM, Adam P. Goucher <apgoucher@gmx.com> wrote:
I forgot to cross out K. Also, GENES = JEANS eliminates J.
_B_D_F_____LM__P_R___V_X_Z
BARMY = BALMY proves L = R (the lallation relation!).
Sent: Friday, September 25, 2015 at 10:14 PM From: "Adam P. Goucher" <apgoucher@gmx.com> To: math-fun@mailman.xmission.com Subject: Re: [math-fun] something linguistic
Given that W is trivial (courtesy of Andy Latto), KNOW = NO establishes the triviality of K.
CELL = SELL ==> C = S SCENT = CENT ==> S is trivial (and therefore so is C)
...
Rarer letters are harder to destroy, but:
CUE = QUEUE ==> UE = QUEUE ==> QUE is trivial
We've already shown that E is an identity, so QU is trivial.
U can be destroyed by FLOWER = FLOUR ==> WE = U, so Q is also trivial.
At the moment, we're left with:
_B_D_FG_IJKLM__P_R___V_XYZ
KNIGHT = NIGHT easily gets rid of K, and we have Y = I = E (from TYRE = TIRE and COMPLIMENT = COMPLEMENT), so those all go:
_B_D_FG__JKLM__P_R___V_X_Z
GNOME = NOME (from elliptic functions) kills the G.
_B_D_F___JKLM__P_R___V_X_Z
Also, FAT = PHAT to deduce F = P, but I can't immediately trivialise either.
Sent: Friday, September 25, 2015 at 9:32 PM From: "Andy Latto" <andy.latto@pobox.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] something linguistic
And I haven't seen the proof that H is trivial, but assuming that, then in those dialects that pronounce initial wh and initial w identically,
WHETHER = WEATHER shows that A is trivial
WHICH = WITCH shows that T is trivial.
DAMN = DAM, so N is trivial
On Fri, Sep 25, 2015 at 4:29 PM, Andy Latto <andy.latto@pobox.com> wrote:
TOO = TO = TWO shows that W and O are trivial
On Fri, Sep 25, 2015 at 4:23 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Michael, do you mean homophones (two words spelled differently that sound alike)? Homonyms are spelled the same.
TYRE = TIRE, so Y=I, if that helps.
On Fri, Sep 25, 2015 at 3:57 PM, Michael Kleber <michael.kleber@gmail.com> wrote:
Brought to mind by Wouter's question is: Consider the group generated by the letters A through Z, with the relations that any two English homonyms are equal. Prove that the groups is trivial.
Wouter's question was to show h is trivial, so 25 letters to go :-).
--Michael
-- Forewarned is worth an octopus in the bush. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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-- Andy.Latto@pobox.com
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See http://projecteuclid.org/download/pdf_1/euclid.em/1062620828 which as I recall proves that the homophony group of English is trivial. Jim Propp On Friday, September 25, 2015, Michael Kleber <michael.kleber@gmail.com> wrote:
Brought to mind by Wouter's question is: Consider the group generated by the letters A through Z, with the relations that any two English homonyms are equal. Prove that the groups is trivial.
Wouter's question was to show h is trivial, so 25 letters to go :-).
--Michael
-- Forewarned is worth an octopus in the bush. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com <javascript:;> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
A quick glance at that paper suggests that it's full of humor. (It, too, showed v = 1 as the last letter to fall.) —Dan
On Sep 25, 2015, at 5:53 PM, James Propp <jamespropp@gmail.com> wrote:
See http://projecteuclid.org/download/pdf_1/euclid.em/1062620828 which as I recall proves that the homophony group of English is trivial.
Jim Propp
On Friday, September 25, 2015, Michael Kleber <michael.kleber@gmail.com> wrote:
Brought to mind by Wouter's question is: Consider the group generated by the letters A through Z, with the relations that any two English homonyms are equal. Prove that the groups is trivial.
Wouter's question was to show h is trivial, so 25 letters to go :-).
--Michael
The two papers presented in parallel are not translations of each other except in a very tenuous Hofstadterian sort of way. The paper written in French proves the result we have been wrestling with here; the one written in English proves the same result for French. I suspect it is simply false for Spanish, which has kept a much closer correspondance between spelling and pronunciation than either English or French have. On Fri, Sep 25, 2015 at 8:57 PM, Dan Asimov <asimov@msri.org> wrote:
A quick glance at that paper suggests that it's full of humor.
(It, too, showed v = 1 as the last letter to fall.)
—Dan
On Sep 25, 2015, at 5:53 PM, James Propp <jamespropp@gmail.com> wrote:
See http://projecteuclid.org/download/pdf_1/euclid.em/1062620828 which as I recall proves that the homophony group of English is trivial.
Jim Propp
On Friday, September 25, 2015, Michael Kleber <michael.kleber@gmail.com> wrote:
Brought to mind by Wouter's question is: Consider the group generated by the letters A through Z, with the relations that any two English homonyms are equal. Prove that the groups is trivial.
Wouter's question was to show h is trivial, so 25 letters to go :-).
--Michael
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And for Devanagari, you probably have a free group.
Sent: Saturday, September 26, 2015 at 2:03 AM From: "Allan Wechsler" <acwacw@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] something linguistic
The two papers presented in parallel are not translations of each other except in a very tenuous Hofstadterian sort of way. The paper written in French proves the result we have been wrestling with here; the one written in English proves the same result for French.
I suspect it is simply false for Spanish, which has kept a much closer correspondance between spelling and pronunciation than either English or French have.
On Fri, Sep 25, 2015 at 8:57 PM, Dan Asimov <asimov@msri.org> wrote:
A quick glance at that paper suggests that it's full of humor.
(It, too, showed v = 1 as the last letter to fall.)
—Dan
On Sep 25, 2015, at 5:53 PM, James Propp <jamespropp@gmail.com> wrote:
See http://projecteuclid.org/download/pdf_1/euclid.em/1062620828 which as I recall proves that the homophony group of English is trivial.
Jim Propp
On Friday, September 25, 2015, Michael Kleber <michael.kleber@gmail.com> wrote:
Brought to mind by Wouter's question is: Consider the group generated by the letters A through Z, with the relations that any two English homonyms are equal. Prove that the groups is trivial.
Wouter's question was to show h is trivial, so 25 letters to go :-).
--Michael
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I sent the link to this paper to a Francophile friend who came back with this evidence that English and French are homophonically isomorphic. Un petit d'un petit S'étonne aux Halles Un petit d'un petit Ah! degrés te fallent Indolent qui ne sort cesse Indolent qui ne se mène Qu'importe un petit d'un petit Tout Gai de Reguennes. Taken from Mots-dHeures-Gousses-dAntin-Manuscript <http://www.amazon.com/Mots-dHeures-Gousses-dAntin-Manuscript/dp/0140057307/ref=sr_1_1?ie=UTF8&qid=1443277437&sr=8-1&keywords=Mots+d%27heures+gousses+rames> by Luis d'Antin van Rooten. I had Google Translate read it for me several times before I began to get it. If you are puzzled see wikipedia.org/wiki/Homophonic_translation <https://en.wikipedia.org/wiki/Homophonic_translation> Edwin Clark On Fri, Sep 25, 2015 at 8:53 PM, James Propp <jamespropp@gmail.com> wrote:
See http://projecteuclid.org/download/pdf_1/euclid.em/1062620828 which as I recall proves that the homophony group of English is trivial.
Jim Propp
On Friday, September 25, 2015, Michael Kleber <michael.kleber@gmail.com> wrote:
Brought to mind by Wouter's question is: Consider the group generated by the letters A through Z, with the relations that any two English homonyms are equal. Prove that the groups is trivial.
Wouter's question was to show h is trivial, so 25 letters to go :-).
--Michael
-- Forewarned is worth an octopus in the bush. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com <javascript:;> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Let H denote the standard helix in R^3, namely the set: H := { (cos(t), sin(t), t) | t in R}. Puzzle: ------- Express R^3 as the disjoint union of subsets each isometric to H, or prove this is impossible. —Dan
Do you mean translates and rotations of H ? Otherwise, a line is isometric to H. -- Gene From: Dan Asimov <asimov@msri.org> To: math-fun <math-fun@mailman.xmission.com> Sent: Saturday, September 26, 2015 11:13 AM Subject: [math-fun] Helix puzzle Let H denote the standard helix in R^3, namely the set: H := { (cos(t), sin(t), t) | t in R}. Puzzle: ------- Express R^3 as the disjoint union of subsets each isometric to H, or prove this is impossible. —Dan
Good point. I mean isometric via any isometry of R^3 (which includes translations and rotations). —Dan
On Sep 26, 2015, at 11:51 AM, Eugene Salamin via math-fun <math-fun@mailman.xmission.com> wrote:
Do you mean translates and rotations of H ? Otherwise, a line is isometric to H.
From Dan Asimov on Saturday, September 26, 2015 11:13 AM:
Let H denote the standard helix in R^3, namely the set:
H := { (cos(t), sin(t), t) | t in R}.
Puzzle: -------
Express R^3 as the disjoint union of subsets each isometric to H, or prove this is impossible.
Isn't it possible to just take the union of all helices of the form: H_(x,y) := {(cos(t) + x, sin(t) + y, t) | t in R}
Sent: Saturday, September 26, 2015 at 7:56 PM From: "Dan Asimov" <asimov@msri.org> To: "Eugene Salamin" <gene_salamin@yahoo.com>, math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Helix puzzle
Good point. I mean isometric via any isometry of R^3 (which includes translations and rotations).
—Dan
On Sep 26, 2015, at 11:51 AM, Eugene Salamin via math-fun <math-fun@mailman.xmission.com> wrote:
Do you mean translates and rotations of H ? Otherwise, a line is isometric to H.
From Dan Asimov on Saturday, September 26, 2015 11:13 AM:
Let H denote the standard helix in R^3, namely the set:
H := { (cos(t), sin(t), t) | t in R}.
Puzzle: -------
Express R^3 as the disjoint union of subsets each isometric to H, or prove this is impossible.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Not only possible but highly probable! —Dan
On Sep 26, 2015, at 12:47 PM, Adam P. Goucher <apgoucher@gmx.com> wrote:
Isn't it possible to just take the union of all helices of the form:
H_(x,y) := {(cos(t) + x, sin(t) + y, t) | t in R}
Sent Saturday, September 26, 2015 at 7:56 PM by Dan Asimov
Good point. I mean isometric via any isometry of R^3 (which includes translations and rotations).
—Dan
On Sep 26, 2015, at 11:51 AM, Eugene Salamin via math-fun <math-fun@mailman.xmission.com> wrote:
Do you mean translates and rotations of H ? Otherwise, a line is isometric to H.
From Dan Asimov on Saturday, September 26, 2015 11:13 AM:
Let H denote the standard helix in R^3, namely the set:
H := { (cos(t), sin(t), t) | t in R}.
Puzzle: -------
Express R^3 as the disjoint union of subsets each isometric to H, or prove this is impossible.
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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participants (17)
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Adam P. Goucher -
Allan Wechsler -
Andy Latto -
Dan Asimov -
Dan Asimov -
Eugene Salamin -
Gareth McCaughan -
James Propp -
Michael Kleber -
Mike Speciner -
Mike Stay -
rkg -
rwg -
Thane Plambeck -
Tom Duff -
W. Edwin Clark -
Wouter Meeussen